What do we mean by derivative of a function? What does it tell?

Question

Taking the derivative of any kind of function is easy but I don't know why we take the derivative? Like

$f(x)=x^2$ has the derivative $2x$,

so what does it mean? I don't know how to define dy/dx? Understanding derivative and Integral in terms of physics becomes hard. Like [here]: (https://physics.stackexchange.com/questions/136139/electric-flux-integral-of-e-with-respect-to-a-is-equal-to-total-electric-fl) edit: I don't think it's a duplicate of other question. I've studied Derivative and Integral in maths but when it comes to physics it needs to have concept in our mind. Like this

Answer 1

The derivative function says how fast the original function is changing at each point. If $f(t)$ is the position of a particle or a rocket ship at each time $t$, then the derivative $f'(t)$ is the speed of the particle or the rocket ship at time $t$.

Consider as an example $f(t) = -5t^2 + 20t$. Suppose this describes the height of a rocket above the ground at time $t$. This curve is a downward-facing parabola with $f(0) = f(4) = 0$ and the peak of the parabola at $f(2) = 20$:

At $t=0$ we have $f(t) = 0$ and the rocket is on the ground. The rocket goes up, quickly at first, then more slowly, until at $t=2$ it stops going up and starts to come down, slowly at first, then more quickly as time goes by, until it hits the ground again at $t=4$.

The derivative of $f$ is the speed of the rocket. $$f'(t) = -10 t+20.$$

The derivative is the blue line in the picture. It represents the upward speed of the rocket at each point. When $t=0$, the derivative has the value $20$, representing a fast upward motion. When $t=1$, the upward speed has decreased to $10$. When $t=2$, the rocket has reached the peak of its flight and has stopped going up and is about to come back down. $f'(t) = 0$, meaning that the rocket has no motion at this instant. Then at $t=3$ the derivative is $-10$, which represents a downward motion, and at $t=4$ when the rocket hits the ground its downward motion is twice as fast, since $f(4) = -20$.

Answer 2

Let's assume that you want to measure the amount of the slope between two points on the function $f(x)$, which are $f(x)$ and $f(x+h)$. This is defined as: $$M(h)=\dfrac{f(x+h)-f(x)}{h}$$. $M$ is actually the slope of the secant line which joins the points $(x,f(x))$ and $(x+h,f(x+h))$. Notice that, when we determine a fixed $x$ value, $M(h)$ is a function of $h$ actually, when you plug in an arbitrary $h$ into it, it returns the slope of the secant line between $(x,f(x))$ and $(x+h,f(x+h))$.

Now, think that you want to measure the slope of the tangent line which only touches $(x,f(x))$. Intuitively, one wishes to plug $h=0$ into $M(h)$ and get the slope eventually. But this is not possible since $M(h)$ is not defined at $h=0$: This causes a division by zero. But still, we can make a very educated guess about the slope of the tangent: If the function $M(h)$ is well behaving around $h=0$, which means that the value of $M(h)$ gets closer and closer to a certain value, say, $L$, as $h$ gets closer to $0$, we say that the limit of $M(h)$ as $h \to 0$ is $L$. More precisely, it is $\lim_{h \to 0} M(h) = L$. If $L$ does really satisfy the definition of limit at $h=0$ then it is the "most natural" value which we can fill into $M(0)$, more technically, it is the value which ensures the continuity of the function $M(h)$ at $h=0$.

For this reason, we can pick $L$ as our slope for the tangent of the function at $(x,f(x))$. More rigorously, it is the value

$$\lim_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$$ and it is derivative of the function $f(x)$ at the point $(x,f(x))$. Note that if this limit does not exist ($M(h)$ is not well behaving around $h=0$), then we say that the derivative does not exist at that $x$.

This is a kind of mixed definition of the derivative, which has both analytic and geometrical interpretation in it. Hope it helps.